在具有给定高度和宽度的矩形中。我应该找到最多1s的正方形并在stdout上打印1s的数量，同样在同一个正方形中，不得超过1s的一半，即：（（＃of 1s）/ 2） ＆gt; =（＃of 2s）。 Square总是至少2x2大。 所以对于输入（前两个数字是高度和宽度）：
6 8
0 0 2 2 2 1 2 1
0 1 2 2 1 0 1 1
0 0 1 0 1 2 0 2
2 1 0 2 2 1 1 1
1 2 1 0 0 0 1 0
1 2 0 1 1 2 1 1
正确的答案是9.（正方形是5x5大，左上角是第二排，第三列）
现在我设法写了一个正确执行此操作的程序，但它太慢了。
所以我要求建议如何编写算法以便解决这个问题：{1}在1秒内（正确答案15）和这个：https://justpaste.it/1cfem在4秒内（正确答案） 556）。
编辑：我忘了在广场上提到我的意思是只有广场的周边（四边）我的代码是这样的： 通过输入中的所有字段迭代，并迭代从该字段开始的所有可能的正方形（从可能的最大正方形开始）。然后我有一些条件，比如我打破迭代时，正方形的可能周长小于我迄今为止在周边等中找到的已经最大的1的数量。此外，当我试图找到从在给定的字段中，我记得前一个广场的上侧和左侧，然后只是递减它（如果有1或2）。
但这还不够，因为像这样的解决方案在1分半钟内解决了第二个输入，我需要在4秒内完成。 代码： 注意：矿物质代表1s，有毒物质代表2s
#include <stdio.h>
#include <stdlib.h>
int maxMinerals;
void traverseforH(const int const *map, const int height, const int width) {
const int h1 = height - 1;
const int w1 = width - 1;
int lineOffset = 0;
for (int startY = 0; startY < h1; startY++) {
int yside = height - startY;
if (!(yside * 2 + (yside - 2)*2 > maxMinerals)) {
break;
}
for (int startX = 0; startX < w1; startX++) {
int xside = width - startX;
if (!(xside * 2 + (xside - 2)*2 > maxMinerals)) {
break;
}
int maxBoundl = width;
int maxBoundm = width;
if (startY + maxBoundm - height - startX > 0) {
maxBoundl = height;
maxBoundm = height;
if (startX - startY > 0) {
maxBoundl = maxBoundl + startY - startX;
} else {
maxBoundm = maxBoundm + startX - startY;
}
} else if (startY - startX > 0) {
maxBoundm = maxBoundm + startY - startX;
maxBoundl = maxBoundm;
maxBoundm = maxBoundm + startX - startY;
} else {
maxBoundl = maxBoundl + startY - startX;
}
int mBw = (maxBoundl - 1) * width;
int toxicsLeftSide = 0;
int mineralsLeftSide = 0;
int toxicsUpSide = 0;
int mineralsUpSide = 0;
int mw;
int lastMinerals = 0;
int toxics = 0;
int sidey = lineOffset + width;
for (int x = startX; x < maxBoundm; x++) {
mw = x + lineOffset;
if (map[mw] == 1) {
mineralsUpSide++;
lastMinerals++;
} else if (map[mw]) {
toxicsUpSide++;
toxics++;
}
mw = x + mBw;
if (map[mw] == 1) {
lastMinerals++;
} else if (map[mw]) {
toxics++;
}
}
for (int y = startY + 1; y < maxBoundl - 1; y++) {
mw = startX + sidey;
if (map[mw] == 1) {
mineralsLeftSide++;
lastMinerals++;
} else if (map[mw]) {
toxicsLeftSide++;
toxics++;
}
mw = maxBoundm - 1 + sidey;
if (map[mw] == 1) {
lastMinerals++;
} else if (map[mw]) {
toxics++;
}
sidey = sidey + width;
}
if (map[startX + mBw] == 1) {
mineralsLeftSide++;
} else if (map[startX + mBw]) {
toxicsLeftSide++;
}
int upsideData [2];
upsideData[0] = mineralsUpSide;
upsideData[1] = toxicsUpSide;
if (!(lastMinerals / 2.0 < toxics) && lastMinerals > maxMinerals) {
maxMinerals = lastMinerals;
}
mBw = mBw - width;
int noOfSquares;
if (xside < yside) {
noOfSquares = xside - 1;
} else {
noOfSquares = yside - 1;
}
for (int k = 1; k < noOfSquares; k++) {
int maxBoundy = maxBoundl - k;
int maxBoundx = maxBoundm - k;
if (!(((maxBoundx - startX)*2 + (maxBoundx - 2 - startX)*2) > maxMinerals)) {
break;
}
sidey = lineOffset + width;
lastMinerals = 0;
toxics = 0;
if (map[maxBoundx + lineOffset] == 1) {
mineralsUpSide--;
} else if (map[maxBoundx + lineOffset]) {
toxicsUpSide--;
}
if (map[startX + mBw + width] == 1) {
mineralsLeftSide--;
} else if (map[startX + mBw + width]) {
toxicsLeftSide--;
}
for (int x = startX + 1; x < maxBoundx; x++) {
mw = x + mBw;
if (map[mw] == 1) {
lastMinerals++;
} else if (map[mw]) {
toxics++;
}
}
for (int y = startY + 1; y < maxBoundy - 1; y++) {
mw = maxBoundx - 1 + sidey;
if (map[mw] == 1) {
lastMinerals++;
} else if (map[mw]) {
toxics++;
}
sidey = sidey + width;
}
int finalMinerals = lastMinerals + mineralsLeftSide + mineralsUpSide;
int finalToxics = toxics + toxicsLeftSide + toxicsUpSide;
if (!(finalMinerals / 2.0 < finalToxics) && finalMinerals > maxMinerals) {
maxMinerals = finalMinerals;
}
mBw = mBw - width;
}
}
lineOffset = lineOffset + width;
}
printf("%d\n", maxMinerals);
}
void traverseforW(int *map, const int height, const int width) {
int h1 = height - 1;
int w1 = width - 1;
int lineOffset = 0;
for (int startY = 0; startY < h1; startY++) {
int yside = height - startY;
if (!(yside * 2 + (yside - 2)*2 > maxMinerals)) {
break;
}
for (int startX = 0; startX < w1; startX++) {
int xside = width - startX;
if (!(xside * 2 + (xside - 2)*2 > maxMinerals)) {
break;
}
int maxBoundl = height;
int maxBoundm = height;
if (startX + maxBoundl - width - startY > 0) {
maxBoundl = width;
maxBoundm = width;
if (startX - startY > 0) {
maxBoundl = maxBoundl + startY - startX;
} else {
maxBoundm = maxBoundm + startX - startY;
}
} else if (startY - startX > 0) {
maxBoundm = maxBoundm + startX - startY;
} else {
maxBoundl = maxBoundl + startX - startY;
maxBoundm = maxBoundl;
maxBoundl = maxBoundl + startY - startX;
}
int mBw = (maxBoundl - 1) * width;
int toxicsLeftSide = 0;
int mineralsLeftSide = 0;
int toxicsUpSide = 0;
int mineralsUpSide = 0;
int mw;
int lastMinerals = 0;
int toxics = 0;
int sidey = lineOffset + width;
for (int x = startX; x < maxBoundm; x++) {
mw = x + lineOffset;
if (map[mw] == 1) {
mineralsUpSide++;
lastMinerals++;
} else if (map[mw]) {
toxicsUpSide++;
toxics++;
}
mw = x + mBw;
if (map[mw] == 1) {
lastMinerals++;
} else if (map[mw]) {
toxics++;
}
}
for (int y = startY + 1; y < maxBoundl - 1; y++) {
mw = startX + sidey;
if (map[mw] == 1) {
mineralsLeftSide++;
lastMinerals++;
} else if (map[mw]) {
toxicsLeftSide++;
toxics++;
}
mw = maxBoundm - 1 + sidey;
if (map[mw] == 1) {
lastMinerals++;
} else if (map[mw]) {
toxics++;
}
sidey = sidey + width;
}
if (map[startX + mBw] == 1) {
mineralsLeftSide++;
} else if (map[startX + mBw]) {
toxicsLeftSide++;
}
if (!(lastMinerals / 2.0 < toxics) && lastMinerals > maxMinerals) {
maxMinerals = lastMinerals;
}
mBw = mBw - width;
int noOfSquares;
if (xside < yside) {
noOfSquares = xside - 1;
} else {
noOfSquares = yside - 1;
}
for (int k = 1; k < noOfSquares; k++) {
int maxBoundy = maxBoundl - k;
int maxBoundx = maxBoundm - k;
if (!(((maxBoundx - startX)*2 + (maxBoundx - 2 - startX)*2) > maxMinerals)) {
break;
}
sidey = lineOffset + width;
lastMinerals = 0;
toxics = 0;
if (map[maxBoundx + lineOffset] == 1) {
mineralsUpSide--;
} else if (map[maxBoundx + lineOffset]) {
toxicsUpSide--;
}
if (map[startX + mBw + width] == 1) {
mineralsLeftSide--;
} else if (map[startX + mBw + width]) {
toxicsLeftSide--;
}
int finalMinerals = mineralsUpSide + mineralsLeftSide;
int finalToxics = toxicsLeftSide + toxicsUpSide;
for (int x = startX + 1; x < maxBoundx; x++) {
mw = x + mBw;
if (map[mw] == 1) {
lastMinerals++;
} else if (map[mw]) {
toxics++;
}
}
for (int y = startY + 1; y < maxBoundy - 1; y++) {
mw = maxBoundx - 1 + sidey;
if (map[mw] == 1) {
lastMinerals++;
} else if (map[mw]) {
toxics++;
}
sidey = sidey + width;
}
finalMinerals += lastMinerals;
finalToxics += toxics;
if (!(finalMinerals / 2.0 < finalToxics) && finalMinerals > maxMinerals) {
maxMinerals = finalMinerals;
}
mBw = mBw - width;
}
}
lineOffset = lineOffset + width;
}
printf("%d\n", maxMinerals);
}
int main() {
char hw[14];
FILE * file = fopen("pub01.in", "r");
char c;
int k = 0;
while ((c = fgetc(file)) != '\n') {
hw[k] = c;
k++;
}
int h, w;
sscanf(hw, "%d %d", &h, &w);
int size = h * w;
int* input = malloc(size * sizeof (int) + 1);
k = 0;
while ((c = fgetc(file)) != EOF) {
if (c == '0' || c == '1' || c == '2') {
input[k] = c - '0';
k++;
}
}
input[k] = '\0';
if (h > w) {
traverseforH(input, h, w);
} else {
traverseforW(input, h, w);
}
return 0;
}
答案 0 :(得分：1)
预处理步骤：
第一个预处理矩阵，使用前缀和方法所有行和列，这样您就可以在O（1）中计算方形周长中的1和2的数量。
到目前为止，您将拥有4个数据结构：rowSumFor1，rowSumFor2，colSumFor1，colSumFor2。例如：rowSumFor1 [i] [j]会告诉我们第0行中1的#s为0和j之间的列索引。
时间复杂度：O（w x h）
完整代码：
#include<stdio.h>
int min(int a,int b){
return (a<=b)?a:b;
}
int max(int a,int b){
return (a>=b)?a:b;
}
// currently hard-coding dimensions for test purposes
// horizontal sums
int rowSumFor1[600][600];
int rowSumFor2[600][600];
// vertical sums
int colSumFor1[600][600];
int colSumFor2[600][600];
int main(){
int w,h;
scanf("%d %d",&h,&w);
for(int row=1;row <= h;row++)for(int col=1;col <= w;col++){
int temp;
scanf("%d",&temp);
// first add previous sum
rowSumFor1[row][col]=rowSumFor1[row][col - 1];
rowSumFor2[row][col]=rowSumFor2[row][col - 1];
colSumFor1[col][row]=colSumFor1[col][row - 1];
colSumFor2[col][row]=colSumFor2[col][row - 1];
if(temp==1){
rowSumFor1[row][col]++;
colSumFor1[col][row]++;
}
else if(temp==2){
rowSumFor2[row][col]++;
colSumFor2[col][row]++;
}
else{
// do nothing
}
}
int result = 0,rowId,colId,mlength;
for(int len=min(w,h); len > 1 ; len-- ) // iteration on possible lengths
{
for(int row=1;row <= (h - len + 1);row++)for(int col=1;col <= (w - len + 1);col++){ // iteration on all co-ordinates as upper-left corner of our square
// Do calculation here for properties and necessary checking constraints for validity of this square
// Note: not checking trivial conditions like boundary conditions in square, you will have to!!
// Beware of over-counting of corners here, one way to avoid is to select indices such that they don't overcount corners
// 4x4 square example for counting
// aaab
// d b
// d b
// dccc
int topEdge1 = rowSumFor1[row][col + len - 2] - rowSumFor1[row][col - 1];
int bottomEdge1 = rowSumFor1[row + len - 1][col + len - 1] - rowSumFor1[row + len - 1][col];
int leftEdge1 = colSumFor1[col][row + len - 1] - colSumFor1[col][row];
int rightEdge1 = colSumFor1[col + len - 1][row + len - 2] - colSumFor1[col + len - 1][row - 1];
int ones= topEdge1 + bottomEdge1 + leftEdge1 + rightEdge1; // # of 1s on perimeter of this square
int topEdge2 = rowSumFor2[row][col + len - 2] - rowSumFor2[row][col-1];
int bottomEdge2 = rowSumFor2[row+len-1][col+len-1] - rowSumFor2[row+len-1][col];
int leftEdge2 = colSumFor2[col][row + len - 1] - colSumFor2[col][row];
int rightEdge2 = colSumFor2[col + len - 1][row + len - 2] - colSumFor2[col + len -1][row - 1];
int twos= topEdge2 + bottomEdge2 + leftEdge2 + rightEdge2; // # of 2s on perimeter of this square
if(ones >= 2* twos){
if(ones > result){
result = ones;
rowId = row;
colId = col;
mlength = len;
}
}
}
}
printf("%d %d %d\n",rowId,colId,mlength);
printf("%d\n",result);
return 0;
}
时间复杂度：O（w x h x min（w，h））
修改
用完整代码替换伪代码。它对OP提出的所有3项测试都有预期效果。